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If there are $D$ inputs from $x_1$ to $x_D$, how can I use a D-D-1 feed forward neural network to implement $\text{XOR}\left(x_1,x_2,\ldots,x_D\right)$ with signum as transformation function?

Here D-D-1 means $D$ inputs (not including $x_0$), one hidden layer with $D$ neurons (not including the constant neuron) and one output layer.

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    $\begingroup$ What are your thoughts? Probably you should start with considering $D = 2$. $\endgroup$ – Random Jack Jan 3 '16 at 15:03
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I can't prove it, but I'd bet that's impossible for $D>2$. It works for $D=2$ as one of the hidden neurons computes $x_1 \land \bar x_2$ and the other computes $\bar x_1 \land x_2$. It works by computing the points where the output should become true (due to symmetries, there are multiple solutions, but all work according to the same principle).

For $D=2$ it only works, as there are $2^{D-1}$ such points. You'd need networks like 3-4-1, 4-8-1, 5-16-1. etc., which sounds pretty unusable.

The problem are the layers. Without them, a 24 neuron network can learn 6-input xor (just tested).

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